Polynomials. Eve Rawley, (EveR) Anne Gloag, (AnneG) Andrew Gloag, (AndrewG)

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1 Polynomials Eve Rawley, (EveR) Anne Gloag, (AnneG) Andrew Gloag, (AndrewG) Say Thanks to the Authors Click (No sign in required)

2 To access a customizable version of this book, as well as other interactive content, visit AUTHORS Eve Rawley, (EveR) Anne Gloag, (AnneG) Andrew Gloag, (AndrewG) CK-12 Foundation is a non-profit organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform. Copyright 2013 CK-12 Foundation, The names CK-12 and CK12 and associated logos and the terms FlexBook and FlexBook Platform (collectively CK-12 Marks ) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License ( licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the CC License ), which is incorporated herein by this reference. Complete terms can be found at Printed: September 23, 2013

3 Chapter 1. Polynomials CHAPTER 1 Polynomials CHAPTER OUTLINE 1.1 Addition and Subtraction of Polynomials 1.2 Multiplication of Polynomials 1.3 Special Products of Polynomials 1.4 Polynomial Equations in Factored Form 1.5 Factoring Quadratic Expressions 1.6 Factoring Special Products 1.7 Factoring Polynomials Completely 1

4 1.1. Addition and Subtraction of Polynomials Addition and Subtraction of Polynomials Learning Objectives Write a polynomial expression in standard form. Classify polynomial expression by degree. Add and subtract polynomials. Solve problems using addition and subtraction of polynomials. Introduction So far we ve seen functions described by straight lines (linear functions) and functions where the variable appeared in the exponent (exponential functions). In this section we ll introduce polynomial functions. A polynomial is made up of different terms that contain positive integer powers of the variables. Here is an example of a polynomial: 4x 3 + 2x 2 3x + 1 Each part of the polynomial that is added or subtracted is called a term of the polynomial. The example above is a polynomial with four terms. The numbers appearing in each term in front of the variable are called the coefficients. The number appearing all by itself without a variable is called a constant. In this case the coefficient of x 3 is 4, the coefficient of x 2 is 2, the coefficient of x is -3 and the constant is 1. Degrees of Polynomials and Standard Form Each term in the polynomial has a different degree. The degree of the term is the power of the variable in that term. 4x 3 has degree 3 and is called a cubic term or 3 rd order term. 2x 2 has degree 2 and is called a quadratic term or 2 nd order term. 3x has degree 1 and is called a linear term or 1 st order term. 1 has degree 0 and is called the constant. 2

5 Chapter 1. Polynomials By definition, the degree of the polynomial is the same as the degree of the term with the highest degree. This example is a polynomial of degree 3, which is also called a cubic polynomial. (Why do you think it is called a cubic?). Polynomials can have more than one variable. Here is another example of a polynomial: t 4 6s 3 t 2 12st + 4s 4 5 This is a polynomial because all the exponents on the variables are positive integers. This polynomial has five terms. Let s look at each term more closely. Note: The degree of a term is the sum of the powers on each variable in the term. In other words, the degree of each term is the number of variables that are multiplied together in that term, whether those variables are the same or different. t 4 6s 3 t 2 12st 4s 4 has a degree of 4, so it s a 4 th order term has a degree of 5, so it s a 5 th order term. has a degree of 2, so it s a 2 nd order term. has a degree of 4, so it s a 4 th order term. 5 is a constant, so its degree is 0. Since the highest degree of a term in this polynomial is 5, then this is polynomial of degree 5 th or a 5 th order polynomial. A polynomial that has only one term has a special name. It is called a monomial (mono means one). A monomial can be a constant, a variable, or a product of a constant and one or more variables. You can see that each term in a polynomial is a monomial, so a polynomial is just the sum of several monomials. Here are some examples of monomials: b 2 2ab x4 29xy Example 1 For the following polynomials, identify the coefficient of each term, the constant, the degree of each term and the degree of the polynomial. a) x 5 3x 3 + 4x 2 5x + 7 b) x 4 3x 3 y 2 + 8x 12 a) x 5 3x 3 + 4x 2 5x + 7 The coefficients of each term in order are 1, -3, 4, and -5 and the constant is 7. The degrees of each term are 5, 3, 2, 1, and 0. Therefore the degree of the polynomial is 5. b) x 4 3x 3 y 2 + 8x 12 The coefficients of each term in order are 1, -3, and 8 and the constant is -12. The degrees of each term are 4, 5, 1, and 0. Therefore the degree of the polynomial is 5. 3

6 1.1. Addition and Subtraction of Polynomials Example 2 Identify the following expressions as polynomials or non-polynomials. a) 5x 5 2x b) 3x 2 2x 2 c) x x 1 d) 5 x 3 +1 e) 4x 1 3 f) 4xy 2 2x 2 y 3 + y 3 3x 3 a) This is a polynomial. b) This is not a polynomial because it has a negative exponent. c) This is not a polynomial because it has a radical. d) This is not a polynomial because the power of x appears in the denominator of a fraction (and there is no way to rewrite it so that it does not). e) This is not a polynomial because it has a fractional exponent. f) This is a polynomial. Often, we arrange the terms in a polynomial in order of decreasing power. This is called standard form. The following polynomials are in standard form: 4x 4 3x 3 + 2x 2 x + 1 a 4 b 3 2a 3 b 3 + 3a 4 b 5ab The first term of a polynomial in standard form is called the leading term, and the coefficient of the leading term is called the leading coefficient. The first polynomial above has the leading term 4x 4, and the leading coefficient is 4. The second polynomial above has the leading term a 4 b 3, and the leading coefficient is 1. Example 3 Rearrange the terms in the following polynomials so that they are in standard form. Indicate the leading term and leading coefficient of each polynomial. a) 7 3x 3 + 4x b) ab a 3 + 2b c) 4b b 2 a) 7 3x 3 + 4x becomes 3x 3 + 4x + 7. Leading term is 3x 3 ; leading coefficient is -3. b) ab a 3 + 2b becomes a 3 + ab + 2b. Leading term is a 3 ; leading coefficient is -1. c) 4b b 2 becomes b 2 4b + 4. Leading term is b 2 ; leading coefficient is 1. 4

7 Chapter 1. Polynomials Simplifying Polynomials A polynomial is simplified if it has no terms that are alike. Like terms are terms in the polynomial that have the same variable(s) with the same exponents, whether they have the same or different coefficients. For example, 2x 2 y and 5x 2 y are like terms, but 6x 2 y and 6xy 2 are not like terms. When a polynomial has like terms, we can simplify it by combining those terms. x 2 + 6xy 4xy + y 2 Like terms We can simplify this polynomial by combining the like terms 6xy and 4xy into (6 4)xy, or 2xy. The new polynomial is x 2 + 2xy + y 2. Example 4 Simplify the following polynomials by collecting like terms and combining them. a) 2x 4x x x b) a 3 b 3 5ab 4 + 2a 3 b a 3 b 3 + 3ab 4 a 2 b a) Rearrange the terms so that like terms are grouped together: ( 4x 2 + x 2 ) + (2x + 4x) + (6 4) Combine each set of like terms: 3x 2 + 6x + 2 b) Rearrange the terms so that like terms are grouped together: (a 3 b 3 a 3 b 3 ) + ( 5ab 4 + 3ab 4 ) + 2a 3 b a 2 b Combine each set of like terms: 0 2ab 4 + 2a 3 b a 2 b = 2ab 4 + 2a 3 b a 2 b Adding and Subtracting Polynomials To add two or more polynomials, write their sum and then simplify by combining like terms. Example 5 Add and simplify the resulting polynomials. a) Add 3x 2 4x + 7 and 2x 3 4x 2 6x + 5 b) Add x 2 2xy + y 2 and 2y 2 3x 2 and 10xy + y 3 a) (3x 2 4x + 7) + (2x 3 4x 2 6x + 5) Group like terms: = 2x 3 + (3x 2 4x 2 ) + ( 4x 6x) + (7 + 5) Simplify: = 2x 3 x 2 10x

8 1.1. Addition and Subtraction of Polynomials b) (x 2 2xy + y 2 ) + (2y 2 3x 2 ) + (10xy + y 3 ) Group like terms: = (x 2 3x 2 ) + (y 2 + 2y 2 ) + ( 2xy + 10xy) + y 3 Simplify: = 2x 2 + 3y 2 + 8xy + y 3 To subtract one polynomial from another, add the opposite of each term of the polynomial you are subtracting. Example 6 a) Subtract x 3 3x 2 + 8x + 12 from 4x 2 + 5x 9 b) Subtract 5b 2 2a 2 from 4a 2 8ab 9b 2 a) b) (4x 2 + 5x 9) (x 3 3x 2 + 8x + 12) = (4x 2 + 5x 9) + ( x 3 + 3x 2 8x 12) Group like terms: = x 3 + (4x 2 + 3x 2 ) + (5x 8x) + ( 9 12) Simplify: = x 3 + 7x 2 3x 21 (4a 2 8ab 9b 2 ) (5b 2 2a 2 ) = (4a 2 8ab 9b 2 ) + ( 5b 2 + 2a 2 ) Group like terms: = (4a 2 + 2a 2 ) + ( 9b 2 5b 2 ) 8ab Simplify: = 6a 2 14b 2 8ab Note: An easy way to check your work after adding or subtracting polynomials is to substitute a convenient value in for the variable, and check that your answer and the problem both give the same value. For example, in part (b) above, if we let a = 2 and b = 3, then we can check as follows: Given (4a 2 8ab 9b 2 ) (5b 2 2a 2 ) 6a 2 14b 2 8ab (4(2) 2 8(2)(3) 9(3) 2 ) (5(3) 2 2(2) 2 ) 6(2) 2 14(3) 2 8(2)(3) (4(4) 8(2)(3) 9(9)) (5(9) 2(4)) 6(4) 14(9) 8(2)(3) ( 113) Since both expressions evaluate to the same number when we substitute in arbitrary values for the variables, we can be reasonably sure that our answer is correct. Note: When you use this method, do not choose 0 or 1 for checking since these can lead to common problems. Problem Solving Using Addition or Subtraction of Polynomials One way we can use polynomials is to find the area of a geometric figure. Example 7 6

9 Chapter 1. Polynomials Write a polynomial that represents the area of each figure shown. a) b) c) d) a) This shape is formed by two squares and two rectangles. The blue square has area y y = y 2. The yellow square has area x x = x 2. The pink rectangles each have area x y = xy. 7

10 1.1. Addition and Subtraction of Polynomials To find the total area of the figure we add all the separate areas: Total area = y 2 + x 2 + xy + xy = y 2 + x 2 + 2xy b) This shape is formed by two squares and one rectangle. The yellow squares each have area a a = a 2. The orange rectangle has area 2a b = 2ab. To find the total area of the figure we add all the separate areas: Total area = a 2 + a 2 + 2ab = 2a 2 + 2ab c) To find the area of the green region we find the area of the big square and subtract the area of the little square. The big square has area : y y = y 2. The little square has area : x x = x 2. Area o f the green region = y 2 x 2 d) To find the area of the figure we can find the area of the big rectangle and add the areas of the pink squares. The pink squares each have area a a = a 2. The blue rectangle has area 3a a = 3a 2. To find the total area of the figure we add all the separate areas: Total area = a 2 + a 2 + a 2 + 3a 2 = 6a 2 Another way to find this area is to find the area of the big square and subtract the areas of the three yellow squares: 8

11 Chapter 1. Polynomials To find the total area of the figure we subtract: The big square has area 3a 3a = 9a 2. The yellow squares each have area a a = a 2. Area = 9a 2 (a 2 + a 2 + a 2 ) = 9a 2 3a 2 = 6a 2 Further Practice For more practice adding and subtracting polynomials, try playing the Battleship game at html. (The problems get harder as you play; watch out for trick questions!) Review Questions Indicate whether each expression is a polynomial. 1. x 2 + 3x x2 y 9y x t2 1 t 2 5. x 2x ( ) 2 6. x 3 2 Express each polynomial in standard form. Give the degree of each polynomial x x + 3x x 5x 2 + 8x x 2 9x x + 2x 2 3x Add and simplify. 12. (x + 8) + ( 3x 5) 13. ( 2x 2 + 4x 12) + (7x + x 2 ) 14. (2a 2 b 2a + 9) + (5a 2 b 4b + 5) 15. (6.9a ( 2 2.3b 2 + 2ab) + (3.1a 2.5b 2 + b) x2 1 4 x + 4) + ( 1 10 x x 2 1 ) 5 Subtract and simplify. 9

12 1.1. Addition and Subtraction of Polynomials ( t + 5t 2 ) (5t 2 + 2t 9) 18. ( y 2 + 4y 5) (5y 2 + 2y + 7) 19. ( 5m 2 m) (3m 2 + 4m 5) 20. (2a 2 b 3ab 2 + 5a 2 b 2 ) (2a 2 b 2 + 4a 2 b 5b 2 ) 21. (3.5x 2 y 6xy + 4x) (1.2x 2 y xy + 2y 3) Find the area of the following figures

13 Chapter 1. Polynomials 1.2 Multiplication of Polynomials Learning Objectives Multiply a polynomial by a monomial. Multiply a polynomial by a binomial. Solve problems using multiplication of polynomials. Introduction Just as we can add and subtract polynomials, we can also multiply them. The Distributive Property and the techniques you ve learned for dealing with exponents will be useful here. Multiplying a Polynomial by a Monomial When multiplying polynomials, we must remember the exponent rules that we learned in the last chapter. Especially important is the product rule: x n x m = x n+m. If the expressions we are multiplying have coefficients and more than one variable, we multiply the coefficients just as we would any number and we apply the product rule on each variable separately. Example 1 Multiply the following monomials. a) (2x 2 )(5x 3 ) b) ( 3y 4 )(2y 2 ) c) (3xy 5 )( 6x 4 y 2 ) d) ( 12a 2 b 3 c 4 )( 3a 2 b 2 ) a) (2x 2 )(5x 3 ) = (2 5) (x 2 x 3 ) = 10x 2+3 = 10x 5 b) ( 3y 4 )(2y 2 ) = ( 3 2) (y 4 y 2 ) = 6y 4+2 = 6y 6 c) (3xy 5 )( 6x 4 y 2 ) = 18x 1+4 y 5+2 = 18x 5 y 7 d) ( 12a 2 b 3 c 4 )( 3a 2 b 2 ) = 36a 2+2 b 3+2 c 4 = 36a 4 b 5 c 4 To multiply a polynomial by a monomial, we have to use the Distributive Property. Remember, that property says that a(b + c) = ab + ac. Example 2 Multiply: 11

14 1.2. Multiplication of Polynomials a) 3(x 2 + 3x 5) b) 4x(3x 2 7) c) 7y(4y 2 2y + 1) a) 3(x 2 + 3x 5) = 3(x 2 ) + 3(3x) 3(5) = 3x 2 + 9x 15 b) 4x(3x 2 7) = (4x)(3x 2 ) + (4x)( 7) = 12x 3 28x c) 7y(4y 2 2y + 1) = ( 7y)(4y 2 ) + ( 7y)( 2y) + ( 7y)(1) = 28y y 2 7y Notice that when we use the Distributive Property, the problem becomes a matter of just multiplying monomials by monomials and adding all the separate parts together. Example 3 Multiply: a) 2x 3 ( 3x 4 + 2x 3 10x 2 + 7x + 9) b) 7a 2 bc 3 (5a 2 3b 2 9c 2 ) a) b) 2x 3 ( 3x 4 + 2x 3 10x 2 + 7x + 9) = (2x 3 )( 3x 4 ) + (2x 3 )(2x 3 ) + (2x 3 )( 10x 2 ) + (2x 3 )(7x) + (2x 3 )(9) = 6x 7 + 4x 6 20x x x 3 7a 2 bc 3 (5a 2 3b 2 9c 2 ) = ( 7a 2 bc 3 )(5a 2 ) + ( 7a 2 bc 3 )( 3b 2 ) + ( 7a 2 bc 3 )( 9c 2 ) = 35a 4 bc a 2 b 3 c a 2 bc 5 Multiplying Two Polynomials Let s start by multiplying two binomials together. A binomial is a polynomial with two terms, so a product of two binomials will take the form (a + b)(c + d). We can still use the Distributive Property here if we do it cleverly. First, let s think of the first set of parentheses as one term. The Distributive Property says that we can multiply that term by c, multiply it by d, and then add those two products together: (a + b)(c + d) = (a + b) c + (a + b) d. We can rewrite this expression as c(a + b) + d(a + b). Now let s look at each half separately. We can apply the distributive property again to each set of parentheses in turn, and that gives us c(a+b) +d(a+b) = ca+cb+da+db. What you should notice is that when multiplying any two polynomials, every term in one polynomial is multiplied by every term in the other polynomial. Example 4 Multiply and simplify: (2x + 1)(x + 3) We must multiply each term in the first polynomial by each term in the second polynomial. Let s try to be systematic to make sure that we get all the products. First, multiply the first term in the first set of parentheses by all the terms in the second set of parentheses. 12

15 Chapter 1. Polynomials Now we re done with the first term. Next we multiply the second term in the first parenthesis by all terms in the second parenthesis and add them to the previous terms. Now we re done with the multiplication and we can simplify: (2x)(x) + (2x)(3) + (1)(x) + (1)(3) = 2x 2 + 6x + x + 3 = 2x 2 + 7x + 3 This way of multiplying polynomials is called in-line multiplication or horizontal multiplication. Another method for multiplying polynomials is to use vertical multiplication, similar to the vertical multiplication you learned with regular numbers. Example 5 Multiply and simplify: a) (4x 5)(x 20) b) (3x 2)(3x + 2) c) (3x 2 + 2x 5)(2x 3) d) (x 2 9)(4x 4 + 5x 2 2) a) With horizontal multiplication this would be (4x 5)(x 20) = (4x)(x) + (4x)( 20) + ( 5)(x) + ( 5)( 20) = 4x 2 80x 5x = 4x 2 85x To do vertical multiplication instead, we arrange the polynomials on top of each other with like terms in the same columns: 4x 5 x 20 80x x 2 5x 4x 2 85x Both techniques result in the same answer: 4x 2 85x+100. We ll use vertical multiplication for the other problems. b) 3x 2 3x + 2 6x 4 9x 2 6x 9x 2 + 0x 4 13

16 1.2. Multiplication of Polynomials The answer is 9x 2 4. c) It s better to place the smaller polynomial on the bottom: 3x 2 + 2x 5 2x 3 9x 2 6x x 3 + 4x 2 10x 6x 3 5x 2 16x + 15 The answer is 6x 3 5x 2 16x d) Set up the multiplication vertically and leave gaps for missing powers of x: 4x 4 + 5x 2 2 x x 4 45x x 6 + 5x 4 2x 2 4x 6 31x 4 47x The answer is 4x 6 31x 4 47x The Khan Academy video at shows how multiplying two binomials together is related to the distributive property. Solve Real-World Problems Using Multiplication of Polynomials In this section, we ll see how multiplication of polynomials is applied to finding the areas and volumes of geometric shapes. Example 6 Find the areas of the following figures: a) b) 14

17 Chapter 1. Polynomials Find the volumes of the following figures: c) d) a) We use the formula for the area of a rectangle: Area = length width. For the big rectangle: Length = b + 3, Width = b + 2 Area = (b + 3)(b + 2) = b 2 + 2b + 3b + 6 = b 2 + 5b + 6 b) We could add up the areas of the blue and orange rectangles, but it s easier to just find the area of the whole big rectangle and subtract the area of the yellow rectangle. 15

18 1.2. Multiplication of Polynomials Area of big rectangle = 20(12) = 240 Area of yellow rectangle = (12 x)(20 2x) = x 20x + 2x 2 = x + 2x 2 = 2x 2 44x The desired area is the difference between the two: Area = 240 (2x 2 44x + 240) = ( 2x x 240) = 240 2x x 240 = 2x x c) The volume of this shape = (area of the base)(height). Area of the base = x(x + 2) = x 2 + 2x Height = 2x + 1 Volume = (x 2 + 2x)(2x + 1) = 2x 3 + x 2 + 4x 2 + 2x = 2x 3 + 5x 2 + 2x d) The volume of this shape = (area of the base)(height). Area of the base = (4a 3)(2a + 1) = 8a 2 + 4a 6a 3 = 8a 2 2a 3 Height = a + 4 Volume = (8a 2 2a 3)(a + 4) Let s multiply using the vertical method: 8a 2 2a 3 a a 2 8a 12 8a 3 2a 2 3a 8a a 2 11a 12 The volume is 8a a 2 11a

19 Chapter 1. Polynomials Review Questions Multiply the following monomials. 1. (2x)( 7x) 2. (10x)(3xy) 3. (4mn)(0.5nm 2 ) 4. ( 5a 2 b)( 12a 3 b 3 ) 5. (3xy 2 z 2 )(15x 2 yz 3 ) Multiply and simplify (8x 10) 7. 2x(4x 5) 8. 9x 3 (3x 2 2x + 7) 9. 3x(2y 2 + y 5) q(3q 2 r + 5r) 11. 3a 2 b(9a 2 4b 2 ) 12. (x 3)(x + 2) 13. (a + b)(a 5) 14. (x + 2)(x 2 3) 15. (a 2 + 2)(3a 2 4) 16. (7x 2)(9x 5) 17. (2x 1)(2x 2 x + 3) 18. (3x + 2)(9x 2 6x + 4) 19. (a 2 + 2a 3)(a 2 3a + 4) 20. 3(x 5)(2x + 7) 21. 5x(x + 4)(2x 3) Find the areas of the following figures. 22. Find the volumes of the following figures

20 1.2. Multiplication of Polynomials

21 Chapter 1. Polynomials 1.3 Special Products of Polynomials Learning Objectives Find the square of a binomial Find the product of binomials using sum and difference formula Solve problems using special products of polynomials Introduction We saw that when we multiply two binomials we need to make sure to multiply each term in the first binomial with each term in the second binomial. Let s look at another example. Multiply two linear binomials (binomials whose degree is 1): (2x + 3)(x + 4) When we multiply, we obtain a quadratic polynomial (one with degree 2) with four terms: 2x 2 + 8x + 3x + 12 The middle terms are like terms and we can combine them. We simplify and get 2x x This is a quadratic, or second-degree, trinomial (polynomial with three terms). You can see that every time we multiply two linear binomials with one variable, we will obtain a quadratic polynomial. In this section we ll talk about some special products of binomials. Find the Square of a Binomial One special binomial product is the square of a binomial. Consider the product (x + 4)(x + 4). Since we are multiplying the same expression by itself, that means we are squaring the expression. (x + 4)(x + 4) is the same as (x + 4) 2. When we multiply it out, we get x 2 + 4x + 4x + 16, which simplifies to x 2 + 8x Notice that the two middle terms the ones we added together to get 8x were the same. Is this a coincidence? In order to find that out, let s square a general linear binomial. (a + b) 2 = (a + b)(a + b) = a 2 + ab + ab + b 2 = a 2 + 2ab + b 2 19

22 1.3. Special Products of Polynomials Sure enough, the middle terms are the same. How about if the expression we square is a difference instead of a sum? (a b) 2 = (a b)(a b) = a 2 ab ab + b 2 = a 2 2ab + b 2 It looks like the middle two terms are the same in general whenever we square a binomial. The general pattern is: to square a binomial, take the square of the first term, add or subtract twice the product of the terms, and add the square of the second term. You should remember these formulas: (a + b) 2 = a 2 + 2ab + b 2 and (a b) 2 = a 2 2ab + b 2 Remember! Raising a polynomial to a power means that we multiply the polynomial by itself however many times the exponent indicates. For instance, (a+b) 2 = (a+b)(a+b). Don t make the common mistake of thinking that (a + b) 2 = a 2 + b 2! To see why that s not true, try substituting numbers for a and b into the equation (for example, a = 4 and b = 3), and you will see that it is not a true statement. The middle term, 2ab, is needed to make the equation work. We can apply the formulas for squaring binomials to any number of problems. Example 1 Square each binomial and simplify. a) (x + 10) 2 b) (2x 3) 2 c) (x 2 + 4) 2 d) (5x 2y) 2 Let s use the square of a binomial formula to multiply each expression. a) (x + 10) 2 If we let a = x and b = 10, then our formula (a + b) 2 = a 2 + 2ab + b 2 becomes (x + 10) 2 = x 2 + 2(x)(10) , which simplifies to x x b) (2x 3) 2 If we let a = 2x and b = 3, then our formula (a b) 2 = a 2 2ab + b 2 becomes (2x 3) 2 = (2x 2 ) 2(2x)(3) + (3) 2, which simplifies to 4x 2 12x + 9. c) (x 2 + 4) 2 If we let a = x 2 and b = 4, then (x 2 + 4) 2 = (x 2 ) 2 + 2(x 2 )(4) + (4) 2 = x 4 + 8x d) (5x 2y) 2 20

23 Chapter 1. Polynomials If we let a = 5x and b = 2y, then (5x 2y) 2 = (5x) 2 2(5x)(2y) + (2y) 2 = 25x 2 20xy + 4y 2 Find the Product of Binomials Using Sum and Difference Patterns Another special binomial product is the product of a sum and a difference of terms. For example, let s multiply the following binomials. (x + 4)(x 4) = x 2 4x + 4x 16 = x 2 16 Notice that the middle terms are opposites of each other, so they cancel out when we collect like terms. This is not a coincidence. This always happens when we multiply a sum and difference of the same terms. In general, (a + b)(a b) = a 2 ab + ab b 2 = a 2 b 2 When multiplying a sum and difference of the same two terms, the middle terms cancel out. We get the square of the first term minus the square of the second term. You should remember this formula. Sum and Difference Formula: (a + b)(a b) = a 2 b 2 Let s apply this formula to a few examples. Example 2 Multiply the following binomials and simplify. a) (x + 3)(x 3) b) (5x + 9)(5x 9) c) (2x 3 + 7)(2x 3 7) d) (4x + 5y)(4x 5y) a) Let a = x and b = 3, then: (a + b)(a b) = a 2 b 2 (x + 3)(x 3) = x = x 2 9 b) Let a = 5x and b = 9, then: 21

24 1.3. Special Products of Polynomials (a + b)(a b) = a 2 b 2 (5x + 9)(5x 9) = (5x) = 25x 2 81 c) Let a = 2x 3 and b = 7, then: (2x 3 + 7)(2x 3 7) = (2x 3 ) 2 (7) 2 = 4x 6 49 d) Let a = 4x and b = 5y, then: (4x + 5y)(4x 5y) = (4x) 2 (5y) 2 = 16x 2 25y 2 Solve Real-World Problems Using Special Products of Polynomials Now let s see how special products of polynomials apply to geometry problems and to mental arithmetic. Example 3 Find the area of the following square: The length of each side is (a + b), so the area is (a + b)(a + b). Notice that this gives a visual explanation of the square of a binomial. The blue square has area a 2, the red square has area b 2, and each rectangle has area ab, so added all together, the area (a + b)(a + b) is equal to a 2 + 2ab + b 2. The next example shows how you can use the special products to do fast mental calculations. Example 4 Use the difference of squares and the binomial square formulas to find the products of the following numbers without using a calculator. a) b)

25 Chapter 1. Polynomials c) 45 2 d) The key to these mental tricks is to rewrite each number as a sum or difference of numbers you know how to square easily. a) Rewrite 43 as (50 7) and 57 as (50 + 7). Then = (50 7)(50 + 7) = (50) 2 (7) 2 = = 2451 b) Rewrite 112 as ( ) and 88 as (100 12). Then = ( )(100 12) = (100) 2 (12) 2 = = 9856 c) 45 2 = (40 + 5) 2 = (40) 2 + 2(40)(5) + (5) 2 = = 2025 d) Rewrite 481 as ( ) and 319 as (400 81). Then = ( )(400 81) = (400) 2 (81) 2 (400) 2 is easy - it equals (81) 2 is not easy to do mentally, so let s rewrite 81 as (81) 2 = (80 + 1) 2 = (80) 2 + 2(80)(1) + (1) 2 = = 6561 Then = (400) 2 (81) 2 = = Review Questions Use the special product rule for squaring binomials to multiply these expressions. 1. (x + 9) 2 2. (3x 7) 2 3. (5x y) 2 4. (2x 3 3) 2 5. (4x 2 + y 2 ) 2 6. (8x 3) 2 7. (2x + 5)(5 + 2x) 8. (xy y) 2 Use the special product of a sum and difference to multiply these expressions. 9. (2x 1)(2x + 1) 10. (x 12)(x + 12) 11. (5a 2b)(5a + 2b) 12. (ab 1)(ab + 1) 13. (z 2 + y)(z 2 y) 14. (2q 3 + r 2 )(2q 3 r 2 ) 15. (7s t)(t + 7s) 16. (x 2 y + xy 2 )(x 2 y xy 2 ) Find the area of the lower right square in the following figure. 23

26 1.3. Special Products of Polynomials Multiply the following numbers using special products

27 Chapter 1. Polynomials 1.4 Polynomial Equations in Factored Form Learning Objectives Use the zero-product property. Find greatest common monomial factors. Solve simple polynomial equations by factoring. Introduction In the last few sections, we learned how to multiply polynomials by using the Distributive Property. All the terms in one polynomial had to be multiplied by all the terms in the other polynomial. In this section, you ll start learning how to do this process in reverse. The reverse of distribution is called factoring. The total area of the figure above can be found in two ways. We could find the areas of all the small rectangles and add them: ab + ac + ad + ae + 2a. Or, we could find the area of the big rectangle all at once. Its width is a and its length is b + c + d + e + 2, so its area is a(b + c + d + e + 2). Since the area of the rectangle is the same no matter what method we use, those two expressions must be equal. ab + ac + ad + ae + 2a = a(b + c + d + e + 2) To turn the right-hand side of this equation into the left-hand side, we would use the distributive property. To turn the left-hand side into the right-hand side, we would need to factor it. Since polynomials can be multiplied just like numbers, they can also be factored just like numbers and we ll see later how this can help us solve problems. 25

28 1.4. Polynomial Equations in Factored Form Find the Greatest Common Monomial Factor You will be learning several factoring methods in the next few sections. In most cases, factoring takes several steps to complete because we want to factor completely. That means that we factor until we can t factor any more. Let s start with the simplest step: finding the greatest monomial factor. When we want to factor, we always look for common monomials first. Consider the following polynomial, written in expanded form: ax + bx + cx + dx A common factor is any factor that appears in all terms of the polynomial; it can be a number, a variable or a combination of numbers and variables. Notice that in our example, the factor x appears in all terms, so it is a common factor. To factor out the x, we write it outside a set of parentheses. Inside the parentheses, we write what s left when we divide each term by x: Let s look at more examples. Example 1 Factor: a) 2x + 8 b) 15x 25 c) 3a + 9b + 6 x(a + b + c + d) a) We see that the factor 2 divides evenly into both terms: 2x + 8 = 2(x) + 2(4) We factor out the 2 by writing it in front of a parenthesis: 2( ) Inside the parenthesis we write what is left of each term when we divide by 2: 2(x + 4) b) We see that the factor of 5 divides evenly into all terms: 15x 25 = 5(3x) 5(5) Factor out the 5 to get: 5(3x 5) c) We see that the factor of 3 divides evenly into all terms: 3a + 9b + 6 = 3(a) + 3(3b) + 3(2) Factor 3 to get: 3(a + 3b + 2) Example 2 Find the greatest common factor: a) a 3 3a 2 + 4a b) 12a 4 5a 3 + 7a 2 a) Notice that the factor a appears in all terms of a 3 3a 2 + 4a, but each term has a raised to a different power. The greatest common factor of all the terms is simply a. So first we rewrite a 3 3a 2 + 4a as a(a 2 ) + a( 3a) + a(4). 26

29 Chapter 1. Polynomials Then we factor out the a to get a(a 2 3a + 4). b) The factor a appears in all the terms, and it s always raised to at least the second power. So the greatest common factor of all the terms is a 2. We rewrite the expression 12a 4 5a 3 + 7a 2 as (12a 2 a 2 ) (5a a 2 ) + (7 a 2 ) Factor out the a 2 to get a 2 (12a 2 5a + 7). Example 3 Factor completely: a) 3ax + 9a b) x 3 y + xy c) 5x 3 y 15x 2 y xy 3 a) Both terms have a common factor of 3, but they also have a common factor of a. It s simplest to factor these both out at once, which gives us 3a(x + 3). b) Both x and y are common factors. When we factor them both out at once, we get xy(x 2 + 1). c) The common factors are 5, x, and y. Factoring out 5xy gives us 5xy(x 2 3xy + 5xy 2 ). Use the Zero-Product Property The most useful thing about factoring is that we can use it to help solve polynomial equations. For example, consider an equation like 2x 2 + 5x 42 = 0. There s no good way to isolate x in this equation, so we can t solve it using any of the techniques we ve already learned. But the left-hand side of the equation can be factored, making the equation (x + 6)(2x 7) = 0. How is this helpful? The answer lies in a useful property of multiplication: if two numbers multiply to zero, then at least one of those numbers must be zero. This is called the Zero-Product Property. What does this mean for our polynomial equation? Since the product equals 0, then at least one of the factors on the left-hand side must equal zero. So we can find the two solutions by setting each factor equal to zero and solving each equation separately. Setting the factors equal to zero gives us: Solving both of those equations gives us: (x + 6) = 0 OR (2x 7) = 0 x + 6 = 0 2x 7 = 0 x = 6 OR 2x = 7 x = 7 2 Notice that the solution is x = 6OR x = 7 2. The OR means that either of these values of x would make the product of the two factors equal to zero. Let s plug the solutions back into the equation and check that this is correct. 27

30 1.4. Polynomial Equations in Factored Form Both solutions check out. Check : x = 6; Check : x = 7 2 (x + 6)(2x 7) = (x + 6)(2x 7) = ( )( 7 ( 6 + 6)(2( 6) 7) = ) 2 7 = ( ) 19 (0)( 19) = 0 (7 7) = 2 ( ) 19 (0) = 0 2 Factoring a polynomial is very useful because the Zero-Product Property allows us to break up the problem into simpler separate steps. When we can t factor a polynomial, the problem becomes harder and we must use other methods that you will learn later. As a last note in this section, keep in mind that the Zero-Product Property only works when a product equals zero. For example, if you multiplied two numbers and the answer was nine, that wouldn t mean that one or both of the numbers must be nine. In order to use the property, the factored polynomial must be equal to zero. Example 4 Solve each equation: a) (x 9)(3x + 4) = 0 b) x(5x 4) = 0 c) 4x(x + 6)(4x 9) = 0 Since all the polynomials are in factored form, we can just set each factor equal to zero and solve the simpler equations separately a) (x 9)(3x + 4) = 0 can be split up into two linear equations: x 9 = 0 3x + 4 = 0 x = 9 or 3x = 4 x = 4 3 b) x(5x 4) = 0 can be split up into two linear equations: 5x 4 = 0 x = 0 or 5x = 4 x = 4 5 c) 4x(x + 6)(4x 9) = 0 can be split up into three linear equations: 28

31 Chapter 1. Polynomials 4x = 0 x + 6 = 0 4x 9 = 0 x = 0 4 or x = 6 or 4x = 9 x = 0 x = 9 4 Solve Simple Polynomial Equations by Factoring Now that we know the basics of factoring, we can solve some simple polynomial equations. We already saw how we can use the Zero-Product Property to solve polynomials in factored form now we can use that knowledge to solve polynomials by factoring them first. Here are the steps: a) If necessary, rewrite the equation in standard form so that the right-hand side equals zero. b) Factor the polynomial completely. c) Use the zero-product rule to set each factor equal to zero. d) Solve each equation from step 3. e) Check your answers by substituting your solutions into the original equation Example 5 Solve the following polynomial equations. a) x 2 2x = 0 b) 2x 2 = 5x c) 9x 2 y 6xy = 0 a) x 2 2x = 0 Rewrite: this is not necessary since the equation is in the correct form. Factor: The common factor is x, so this factors as x(x 2) = 0. Set each factor equal to zero: Solve: x = 0 or x 2 = 0 x = 0 or x = 2 Check: Substitute each solution back into the original equation. x = 0 (0) 2 2(0) = 0 x = 2 (2) 2 2(2) = 4 4 = 0 works out works out 29

32 1.4. Polynomial Equations in Factored Form Answer: x = 0,x = 2 b) 2x 2 = 5x Rewrite: 2x 2 = 5x 2x 2 5x = 0 Factor: The common factor is x, so this factors as x(2x 5) = 0. Set each factor equal to zero: Solve: x = 0 or 2x 5 = 0 x = 0 or 2x = 5 Check: Substitute each solution back into the original equation. x = 5 2 x = 0 2(0) 2 = 5(0) 0 = 0 x = 5 ( ) = = = 25 2 works out works out Answer: x = 0,x = 5 2 c) 9x 2 y 6xy = 0 Rewrite: not necessary Factor: The common factor is 3xy, so this factors as 3xy(3x 2) = 0. Set each factor equal to zero: 3 = 0 is never true, so this part does not give a solution. The factors we have left give us: Solve: x = 0 or y = 0 or 3x 2 = 0 x = 0 or y = 0 or 3x = 2 x = 2 3 Check: Substitute each solution back into the original equation. x = 0 9(0)y 6(0)y = 0 0 = 0 works out y = 0 9x 2 (0) 6x(0) = 0 0 = 0 works out x = 2 ( ) y y = 9 4 y 4y = 4y 4y = 0 9 works out Answer: x = 0,y = 0,x =

33 Chapter 1. Polynomials Review Questions Factor out the greatest common factor in the following polynomials. 1. 2x 2 5x 2. 3x 3 21x 3. 5x x x x 2 2x 5. 10x x 5 4x xy + 24xy xy a 3 7a 8. 3y + 6z 9. 10a 3 4ab y y xy 2 z + 4x 3 y 12. 2a 4a xy 2 10xy + 5y 2 Solve the following polynomial equations. 14. x(x + 12) = (2x + 1)(2x 1) = (x 5)(2x + 7)(3x 4) = x(x + 9)(7x 20) = x(3 + y) = x(x 2y) = y 3y 2 = x 2 = 27x 22. 4a 2 + a = b b = x 2 = x 3 5x 2 = 0 31

34 1.5. Factoring Quadratic Expressions Factoring Quadratic Expressions Learning Objectives Write quadratic equations in standard form. Factor quadratic expressions for different coefficient values. Write Quadratic Expressions in Standard Form Quadratic polynomials are polynomials of the 2 nd degree. The standard form of a quadratic polynomial is written as ax 2 + bx + c where a,b, and c stand for constant numbers. Factoring these polynomials depends on the values of these constants. In this section we ll learn how to factor quadratic polynomials for different values of a,b, and c. (When none of the coefficients are zero, these expressions are also called quadratic trinomials, since they are polynomials with three terms.) You ve already learned how to factor quadratic polynomials where c = 0. For example, for the quadratic ax 2 + bx, the common factor is x and this expression is factored as x(ax + b). Now we ll see how to factor quadratics where c is nonzero. Factor when a = 1, b is Positive, and c is Positive First, let s consider the case where a = 1,b is positive and c is positive. The quadratic trinomials will take the form x 2 + bx + c You know from multiplying binomials that when you multiply two factors (x + m)(x + n), you get a quadratic polynomial. Let s look at this process in more detail. First we use distribution: (x + m)(x + n) = x 2 + nx + mx + mn Then we simplify by combining the like terms in the middle. We get: (x + m)(x + n) = x 2 + (n + m)x + mn 32

35 Chapter 1. Polynomials So to factor a quadratic, we just need to do this process in reverse. We see that is the same form as x 2 + (n + m)x + mn x 2 + bx + c This means that we need to find two numbers m and n where n + m = b and mn = c The factors of x 2 + bx + c are always two binomials (x + m)(x + n) such that n + m = b and mn = c. Example 1 Factor x 2 + 5x + 6. We are looking for an answer that is a product of two binomials in parentheses: (x )(x ) We want two numbers m and n that multiply to 6 and add up to 5. A good strategy is to list the possible ways we can multiply two numbers to get 6 and then see which of these numbers add up to 5: 6 = 1 6 and = 7 6 = 2 3 and = 5 T his is the correct choice. So the answer is (x + 2)(x + 3). We can check to see if this is correct by multiplying (x + 2)(x + 3): x + 2 x + 3 3x + 6 x 2 + 2x x 2 + 5x + 6 The answer checks out. Example 2 Factor x 2 + 7x

36 1.5. Factoring Quadratic Expressions We are looking for an answer that is a product of two binomials in parentheses: (x )(x ) The number 12 can be written as the product of the following numbers: 12 = 1 12 and = = 2 6 and = 8 12 = 3 4 and = 7 T his is the correct choice. The answer is (x + 3)(x + 4). Example 3 Factor x 2 + 8x We are looking for an answer that is a product of two binomials in parentheses: (x )(x ) The number 12 can be written as the product of the following numbers: 12 = 1 12 and = = 2 6 and = 8 T his is the correct choice. 12 = 3 4 and = 7 The answer is (x + 2)(x + 6). Example 4 Factor x x We are looking for an answer that is a product of two binomials in parentheses: (x )(x ) The number 36 can be written as the product of the following numbers: 36 = 1 36 and = = 2 18 and = = 3 12 and = = 4 9 and = = 6 6 and = 12 T his is the correct choice. The answer is (x + 6)(x + 6). Factor when a = 1, b is Negative and c is Positive Now let s see how this method works if the middle coefficient is negative. Example 5 34

37 Chapter 1. Polynomials Factor x 2 6x + 8. We are looking for an answer that is a product of two binomials in parentheses: (x )(x ) When negative coefficients are involved, we have to remember that negative factors may be involved also. The number 8 can be written as the product of the following numbers: but also 8 = 1 8 and = 9 and 8 = ( 1) ( 8) and 1 + ( 8) = 9 but also 8 = 2 4 and = 6 8 = ( 2) ( 4) and 2 + ( 4) = 6 T his is the correct choice. The answer is (x 2)(x 4). We can check to see if this is correct by multiplying (x 2)(x 4): x 2 x 4 4x + 8 x 2 2x x 2 6x + 8 The answer checks out. Example 6 Factor x 2 17x We are looking for an answer that is a product of two binomials in parentheses: (x )(x ) The number 16 can be written as the product of the following numbers: 16 = 1 16 and = = ( 1) ( 16) and 1 + ( 16) = 17 T his is the correct choice. 16 = 2 8 and = = ( 2) ( 8) and 2 + ( 8) = = 4 4 and = 8 16 = ( 4) ( 4) and 4 + ( 4) = 8 35

38 1.5. Factoring Quadratic Expressions The answer is (x 1)(x 16). In general, whenever b is negative and a and c are positive, the two binomial factors will have minus signs instead of plus signs. Factor when a = 1 and c is Negative Now let s see how this method works if the constant term is negative. Example 7 Factor x 2 + 2x 15. We are looking for an answer that is a product of two binomials in parentheses: (x )(x ) Once again, we must take the negative sign into account. The number -15 can be written as the product of the following numbers: 15 = 1 15 and = = 1 ( 15) and 1 + ( 15) = = 3 5 and = 2 T his is the correct choice. 15 = 3 ( 5) and 3 + ( 5) = 2 The answer is (x 3)(x + 5). We can check to see if this is correct by multiplying: x 3 x + 5 5x 15 x 2 3x x 2 + 2x 15 The answer checks out. Example 8 Factor x 2 10x 24. We are looking for an answer that is a product of two binomials in parentheses: (x )(x ) The number -24 can be written as the product of the following numbers: 36

39 Chapter 1. Polynomials 24 = 1 24 and = = 1 ( 24) and 1 + ( 24) = = 2 12 and = = 2 ( 12) and 2 + ( 12) = 10 T his is the correct choice. 24 = 3 8 and = 5 24 = 3 ( 8) and 3 + ( 8) = 5 24 = 4 6 and = 2 24 = 4 ( 6) and 4 + ( 6) = 2 The answer is (x 12)(x + 2). Example 9 Factor x x 35. We are looking for an answer that is a product of two binomials in parentheses: (x )(x ) The number -35 can be written as the product of the following numbers: 35 = 1 35 and = 34 T his is the correct choice. 35 = 1 ( 35) and 1 + ( 35) = = 5 7 and = 2 35 = 5 ( 7) and 5 + ( 7) = 2 The answer is (x 1)(x + 35). Factor when a = - 1 When a = 1, the best strategy is to factor the common factor of -1 from all the terms in the quadratic polynomial and then apply the methods you learned so far in this section Example 10 Factor x 2 + x + 6. First factor the common factor of -1 from each term in the trinomial. Factoring -1 just changes the signs of each term in the expression: x 2 + x + 6 = (x 2 x 6) We re looking for a product of two binomials in parentheses: (x )(x ) Now our job is to factor x 2 x 6. The number -6 can be written as the product of the following numbers: 37

40 1.5. Factoring Quadratic Expressions 6 = 1 6 and = 5 6 = 1 ( 6) and 1 + ( 6) = 5 6 = 2 3 and = 1 6 = 2 ( 3) and 2 + ( 3) = 1 T his is the correct choice. The answer is (x 3)(x + 2). Lesson Summary A quadratic of the form x 2 + bx + c factors as a product of two binomials in parentheses: (x + m)(x + n) If b and c are positive, then both m and n are positive. Example: x 2 + 8x + 12 factors as (x + 6)(x + 2). If b is negative and c is positive, then both m and n are negative. Example: x 2 6x + 8 factors as (x 2)(x 4). If c is negative, then either m is positive and n is negative or vice-versa. Example: x 2 + 2x 15 factors as (x + 5)(x 3). Example: x x 35 factors as (x + 35)(x 1). If a = 1, factor out -1 from each term in the trinomial and then factor as usual. The answer will have the form: (x + m)(x + n) Example: x 2 + x + 6 factors as (x 3)(x + 2). Review Questions Factor the following quadratic polynomials x x x x x x x x x 2 11x x 2 13x x 2 14x x 2 9x x 2 + 5x x 2 + 6x x 2 + 7x 78

41 Chapter 1. Polynomials 12. x 2 + 4x x 2 12x x 2 5x x 2 3x x 2 x x 2 2x x 2 5x x x x x x x x x x x x 2 17x x 2 + 5x 36 39

42 1.6. Factoring Special Products Factoring Special Products Learning Objectives Factor the difference of two squares. Factor perfect square trinomials. Solve quadratic polynomial equation by factoring. Introduction When you learned how to multiply binomials we talked about two special products. The sum and difference formula: (a + b)(a b) = a 2 b 2 The square of a binomial formulas: (a + b) 2 = a 2 + 2ab + b 2 (a b) 2 = a 2 2ab + b 2 In this section we ll learn how to recognize and factor these special products. Factor the Difference of Two Squares We use the sum and difference formula to factor a difference of two squares. A difference of two squares is any quadratic polynomial in the form a 2 b 2, where a and b can be variables, constants, or just about anything else. The factors of a 2 b 2 are always (a + b)(a b); the key is figuring out what the a and b terms are. Example 1 Factor the difference of squares: a) x 2 9 b) x c) x 2 1 a) Rewrite x 2 9 as x Now it is obvious that it is a difference of squares. The difference of squares formula is: a 2 b 2 = (a + b)(a b) Let s see how our problem matches with the formula: x = (x + 3)(x 3) The answer is: x 2 9 = (x + 3)(x 3) 40

43 Chapter 1. Polynomials We can check to see if this is correct by multiplying (x + 3)(x 3): The answer checks out. x + 3 x 3 3x 9 x 2 + 3x x 2 + 0x 9 Note: We could factor this polynomial without recognizing it as a difference of squares. With the methods we learned in the last section we know that a quadratic polynomial factors into the product of two binomials: (x )(x ) We need to find two numbers that multiply to -9 and add to 0 (since there is no x term, that s the same as if the x term had a coefficient of 0). We can write -9 as the following products: 9 = 1 9 and = 8 9 = 1 ( 9) and 1 + ( 9) = 8 9 = 3 ( 3) and 3 + ( 3) = 0 T hese are the correct numbers. We can factor x 2 9 as (x +3)(x 3), which is the same answer as before. You can always factor using the methods you learned in the previous section, but recognizing special products helps you factor them faster. b) Rewrite x as x This factors as (x + 10)(x 10). c) Rewrite x 2 1 as x This factors as (x + 1)(x 1). Example 2 Factor the difference of squares: a) 16x 2 25 b) 4x 2 81 c) 49x 2 64 a) Rewrite 16x 2 25 as (4x) This factors as (4x + 5)(4x 5). b) Rewrite 4x 2 81 as (2x) This factors as (2x + 9)(2x 9). c) Rewrite 49x 2 64 as (7x) This factors as (7x + 8)(7x 8). Example 3 Factor the difference of squares: a) x 2 y 2 b) 9x 2 4y 2 c) x 2 y

44 1.6. Factoring Special Products a) x 2 y 2 factors as (x + y)(x y). b) Rewrite 9x 2 4y 2 as (3x) 2 (2y) 2. This factors as (3x + 2y)(3x 2y). c) Rewrite x 2 y 2 1 as (xy) This factors as (xy + 1)(xy 1). Example 4 Factor the difference of squares: a) x 4 25 b) 16x 4 y 2 c) x 2 y 8 64z 2 a) Rewrite x 4 25 as (x 2 ) This factors as (x 2 + 5)(x 2 5). b) Rewrite 16x 4 y 2 as (4x 2 ) 2 y 2. This factors as (4x 2 + y)(4x 2 y). c) Rewrite x 2 y 4 64z 2 as (xy 2 ) 2 (8z) 2. This factors as (xy 2 + 8z)(xy 2 8z). Factor Perfect Square Trinomials We use the square of a binomial formula to factor perfect square trinomials. A perfect square trinomial has the form a 2 + 2ab + b 2 or a 2 2ab + b 2. In these special kinds of trinomials, the first and last terms are perfect squares and the middle term is twice the product of the square roots of the first and last terms. In a case like this, the polynomial factors into perfect squares: a 2 + 2ab + b 2 = (a + b) 2 a 2 2ab + b 2 = (a b) 2 Once again, the key is figuring out what the a and b terms are. Example 5 Factor the following perfect square trinomials: a) x 2 + 8x + 16 b) x 2 4x + 4 c) x x + 49 a) The first step is to recognize that this expression is a perfect square trinomial. First, we can see that the first term and the last term are perfect squares. We can rewrite x 2 + 8x + 16 as x 2 + 8x Next, we check that the middle term is twice the product of the square roots of the first and the last terms. This is true also since we can rewrite x 2 + 8x + 16 as x x This means we can factor x 2 + 8x + 16 as (x + 4) 2. We can check to see if this is correct by multiplying (x + 4) 2 = (x + 4)(x + 4) : 42

45 Chapter 1. Polynomials x + 4 x + 4 4x + 16 x 2 + 4x x 2 + 8x + 16 The answer checks out. Note: We could factor this trinomial without recognizing it as a perfect square. We know that a trinomial factors as a product of two binomials: (x )(x ) We need to find two numbers that multiply to 16 and add to 8. We can write 16 as the following products: 16 = 1 16 and = = 2 8 and = = 4 4 and = 8 T hese are the correct numbers So we can factor x 2 + 8x + 16 as (x + 4)(x + 4), which is the same as (x + 4) 2. Once again, you can factor perfect square trinomials the normal way, but recognizing them as perfect squares gives you a useful shortcut. b) Rewrite x 2 + 4x + 4 as x ( 2) x + ( 2) 2. We notice that this is a perfect square trinomial, so we can factor it as (x 2) 2. c) Rewrite x x + 49 as x x We notice that this is a perfect square trinomial, so we can factor it as (x + 7) 2. Example 6 Factor the following perfect square trinomials: a) 4x x + 25 b) 9x 2 24x + 16 c) x 2 + 2xy + y 2 a) Rewrite 4x x + 25 as (2x) (2x) We notice that this is a perfect square trinomial and we can factor it as (2x + 5) 2. b) Rewrite 9x 2 24x + 16 as (3x) ( 4) (3x) + ( 4) 2. We notice that this is a perfect square trinomial and we can factor it as (3x 4) 2. We can check to see if this is correct by multiplying (3x 4) 2 = (3x 4)(3x 4): 43

46 1.6. Factoring Special Products The answer checks out. c) x 2 + 2xy + y 2 3x 4 3x 4 12x x 2 12x 9x 2 24x + 16 We notice that this is a perfect square trinomial and we can factor it as (x + y) 2. For more examples of factoring perfect square trinomials, watch the videos at m/perfect-square-trinomial.html. Solve Quadratic Polynomial Equations by Factoring With the methods we ve learned in the last two sections, we can factor many kinds of quadratic polynomials. This is very helpful when we want to solve them. Remember the process we learned earlier: 1. If necessary, rewrite the equation in standard form so that the right-hand side equals zero. 2. Factor the polynomial completely. 3. Use the zero-product rule to set each factor equal to zero. 4. Solve each equation from step Check your answers by substituting your solutions into the original equation We can use this process to solve quadratic polynomials using the factoring methods we just learned. Example 7 Solve the following polynomial equations. a) x 2 + 7x + 6 = 0 b) x 2 8x = 12 c) x 2 = 2x + 15 a) Rewrite: We can skip this since the equation is in the correct form already. Factor: We can write 6 as a product of the following numbers: 6 = 1 6 and = 7 T his is the correct choice. 6 = 2 3 and = 5 x 2 + 7x + 6 = 0 factors as (x + 1)(x + 6) = 0. Set each factor equal to zero: x + 1 = 0 or x + 6 = 0 44

47 Chapter 1. Polynomials Solve: x = 1 or x = 6 Check: Substitute each solution back into the original equation. x = 1 ( 1) 2 + 7( 1) + 6 = = 0 checks out x = 6 ( 6) 2 + 7( 6) + 6 = = 0 checks out b) Rewrite: x 2 8x = 12 is rewritten as x 2 8x + 12 = 0 Factor: We can write 12 as a product of the following numbers: 12 = 1 12 and = = 1 ( 12) and 1 + ( 12) = = 2 6 and = 8 12 = 2 ( 6) and 2 + ( 6) = 8 T his is the correct choice. 12 = 3 4 and = 7 12 = 3 ( 4) and 3 + ( 4) = 7 x 2 + 8x + 12 = 0 factors as (x 2)(x 6) = 0. Set each factor equal to zero: Solve: x 2 = 0 or x 6 = 0 x = 2 or x = 6 Check: Substitute each solution back into the original equation. x = 2 (2) 2 8(2) = 4 16 = 12 checks out x = 6 (6) 2 8(6) = = 12 checks out c) Rewrite: x 2 = 2x + 15 is rewritten as x 2 2x 15 = 0 Factor: We can write -15 as a product of the following numbers: 15 = 1 ( 15) and 1 + ( 15) = = 1 (15) and 1 + (15) = = 3 5 and = 2 15 = 3 ( 5) and 3 + ( 5) = 2 T his is the correct choice. 45

48 1.6. Factoring Special Products x 2 2x 15 = 0 factors as (x + 3)(x 5) = 0 Set each factor equal to zero: Solve: x + 3 = 0 or x 5 = 0 x = 3 or x = 5 Check: Substitute each solution back into the original equation. x = 3 ( 3) 2 = 2( 3) = 9 checks out x = 5 (5) 2 = 2(5) = 25 checks out Example 8 Solve the following polynomial equations: a) x 2 12x + 36 = 0 b) x 2 81 = 0 c) x x = 0 a) x 2 12x + 36 = 0 Rewrite: The equation is in the correct form already. Factor: Rewrite x 2 12x + 36 = 0 as x 2 2 ( 6)x + ( 6) 2. We recognize this as a perfect square. This factors as (x 6) 2 = 0 or (x 6)(x 6) = 0 Set each factor equal to zero: Solve: x 6 = 0 or x 6 = 0 x = 6 or x = 6 Notice that for a perfect square the two solutions are the same. This is called a double root. Check: Substitute each solution back into the original equation. b) x 2 81 = 0 46 x = (6) + 36 = = 0 checks out

49 Chapter 1. Polynomials Rewrite: this is not necessary since the equation is in the correct form already Factor: Rewrite x 2 81 as x We recognize this as a difference of squares. This factors as (x 9)(x + 9) = 0. Set each factor equal to zero: Solve: x 9 = 0 or x + 9 = 0 x = 9 or x = 9 Check: Substitute each solution back into the original equation. x = = = 0 checks out x = 9 ( 9) 2 81 = = 0 checks out c) x x = 0 Rewrite: this is not necessary since the equation is in the correct form already Factor: Rewrite x x as x x We recognize this as a perfect square. This factors as (x + 10) 2 = 0 or (x + 10)(x + 10) = 0 Set each factor equal to zero: Solve: x + 10 = 0 or x + 10 = 0 x = 10 or x = 10 This is a double root. Check: Substitute each solution back into the original equation. x = 10 ( 10) ( 10) = = 0 checks out Review Questions Factor the following perfect square trinomials. 1. x 2 + 8x x 2 18x

50 1.6. Factoring Special Products 3. x x x x x 2 4x x x x 2 12xy + 9y 2 8. x x Factor the following differences of squares. 9. x x x x x x x x 2 y x 2 81y 2 Solve the following quadratic equations using factoring. 18. x 2 11x + 30 = x 2 + 4x = x = 14x 21. x 2 64 = x 2 24x = x 2 25 = x x = x 2 16x 60 = 0 48

51 Chapter 1. Polynomials 1.7 Factoring Polynomials Completely Learning Objectives Factor out a common binomial. Factor by grouping. Factor a quadratic trinomial where a 1. Solve real world problems using polynomial equations. Introduction We say that a polynomial is factored completely when we can t factor it any more. Here are some suggestions that you should follow to make sure that you factor completely: Factor all common monomials first. Identify special products such as difference of squares or the square of a binomial. Factor according to their formulas. If there are no special products, factor using the methods we learned in the previous sections. Look at each factor and see if any of these can be factored further. Example 1 Factor the following polynomials completely. a) 6x 2 30x + 24 b) 2x 2 8 c) x 3 + 6x 2 + 9x a) Factor out the common monomial. In this case 6 can be divided from each term: 6(x 2 5x 6) There are no special products. We factor x 2 5x + 6 as a product of two binomials: (x )(x ) The two numbers that multiply to 6 and add to -5 are -2 and -3, so: 6(x 2 5x + 6) = 6(x 2)(x 3) If we look at each factor we see that we can factor no more. The answer is 6(x 2)(x 3). 49

52 1.7. Factoring Polynomials Completely b) Factor out common monomials: 2x 2 8 = 2(x 2 4) We recognize x 2 4 as a difference of squares. We factor it as (x + 2)(x 2). If we look at each factor we see that we can factor no more. The answer is 2(x + 2)(x 2). c) Factor out common monomials: x 3 + 6x 2 + 9x = x(x 2 + 6x + 9) We recognize x 2 + 6x + 9 as a perfect square and factor it as (x + 3) 2. If we look at each factor we see that we can factor no more. The answer is x(x + 3) 2. Example 2 Factor the following polynomials completely: a) 2x b) x 5 8x x a) Factor out the common monomial. In this case, factor out -2 rather than 2. (It s always easier to factor out the negative number so that the highest degree term is positive.) 2x = 2(x 4 81) We recognize expression in parenthesis as a difference of squares. We factor and get: 2(x 2 9)(x 2 + 9) If we look at each factor we see that the first parenthesis is a difference of squares. We factor and get: 2(x + 3)(x 3)(x 2 + 9) If we look at each factor now we see that we can factor no more. The answer is 2(x + 3)(x 3)(x 2 + 9). b) Factor out the common monomial: x 5 8x x = x(x 4 8x ) We recognize x 4 8x as a perfect square and we factor it as x(x 2 4) 2. We look at each term and recognize that the term in parentheses is a difference of squares. We factor it and get ((x + 2)(x 2)) 2, which we can rewrite as (x + 2) 2 (x 2) 2. If we look at each factor now we see that we can factor no more. The final answer is x(x + 2) 2 (x 2) 2. Factor out a Common Binomial The first step in the factoring process is often factoring out the common monomials from a polynomial. But sometimes polynomials have common terms that are binomials. For example, consider the following expression: 50

53 Chapter 1. Polynomials x(3x + 2) 5(3x + 2) Since the term (3x + 2) appears in both terms of the polynomial, we can factor it out. We write that term in front of a set of parentheses containing the terms that are left over: (3x + 2)(x 5) This expression is now completely factored. Let s look at some more examples. Example 3 Factor out the common binomials. a) 3x(x 1) + 4(x 1) b) x(4x + 5) + (4x + 5) a) 3x(x 1) + 4(x 1) has a common binomial of (x 1). When we factor out the common binomial we get (x 1)(3x + 4). b) x(4x + 5) + (4x + 5) has a common binomial of (4x + 5). When we factor out the common binomial we get (4x + 5)(x + 1). Factor by Grouping Sometimes, we can factor a polynomial containing four or more terms by factoring common monomials from groups of terms. This method is called factor by grouping. The next example illustrates how this process works. Example 4 Factor 2x + 2y + ax + ay. There is no factor common to all the terms. However, the first two terms have a common factor of 2 and the last two terms have a common factor of a. Factor 2 from the first two terms and factor a from the last two terms: 2x + 2y + ax + ay = 2(x + y) + a(x + y) Now we notice that the binomial (x + y) is common to both terms. We factor the common binomial and get: (x + y)(2 + a) Example 5 51

54 1.7. Factoring Polynomials Completely Factor 3x 2 + 6x + 4x + 8. We factor 3 from the first two terms and factor 4 from the last two terms: 3x(x + 2) + 4(x + 2) Now factor (x + 2) from both terms: (x + 2)(3x + 4). Now the polynomial is factored completely. Factor Quadratic Trinomials Where a 1 Factoring by grouping is a very useful method for factoring quadratic trinomials of the form ax 2 + bx + c, where a 1. A quadratic like this doesn t factor as (x ± m)(x ± n), so it s not as simple as looking for two numbers that multiply to c and add up to b. Instead, we also have to take into account the coefficient in the first term. To factor a quadratic polynomial where a 1, we follow these steps: 1. We find the product ac. 2. We look for two numbers that multiply to ac and add up to b. 3. We rewrite the middle term using the two numbers we just found. 4. We factor the expression by grouping. Let s apply this method to the following examples. Example 6 Factor the following quadratic trinomials by grouping. a) 3x 2 + 8x + 4 b) 6x 2 11x + 4 c) 5x 2 6x + 1 Let s follow the steps outlined above: a) 3x 2 + 8x + 4 Step 1: ac = 3 4 = 12 Step 2: The number 12 can be written as a product of two numbers in any of these ways: 12 = 1 12 and = = 2 6 and = 8 T his is the correct choice. 12 = 3 4 and = 7 Step 3: Re-write the middle term: 8x = 2x + 6x, so the problem becomes: 52

55 Chapter 1. Polynomials 3x 2 + 8x + 4 = 3x 2 + 2x + 6x + 4 Step 4: Factor an x from the first two terms and a 2 from the last two terms: Now factor the common binomial (3x + 2): x(3x + 2) + 2(3x + 2) To check if this is correct we multiply (3x + 2)(x + 2): (3x + 2)(x + 2) T his is the answer. 3x + 2 x + 2 6x + 4 3x 2 + 2x 3x 2 + 8x + 4 The solution checks out. b) 6x 2 11x + 4 Step 1: ac = 6 4 = 24 Step 2: The number 24 can be written as a product of two numbers in any of these ways: 24 = 1 24 and = = 1 ( 24) and 1 + ( 24) = = 2 12 and = = 2 ( 12) and 2 + ( 12) = = 3 8 and = = 3 ( 8) and 3 + ( 8) = 11 T his is the correct choice. 24 = 4 6 and = = 4 ( 6) and 4 + ( 6) = 10 Step 3: Re-write the middle term: 11x = 3x 8x, so the problem becomes: 6x 2 11x + 4 = 6x 2 3x 8x + 4 Step 4: Factor by grouping: factor a 3x from the first two terms and a -4 from the last two terms: 3x(2x 1) 4(2x 1) 53

56 1.7. Factoring Polynomials Completely Now factor the common binomial (2x 1): (2x 1)(3x 4) T his is the answer. c) 5x 2 6x + 1 Step 1: ac = 5 1 = 5 Step 2: The number 5 can be written as a product of two numbers in any of these ways: 5 = 1 5 and = 6 5 = 1 ( 5) and 1 + ( 5) = 6 T his is the correct choice. Step 3: Re-write the middle term: 6x = x 5x, so the problem becomes: 5x 2 6x + 1 = 5x 2 x 5x + 1 Step 4: Factor by grouping: factor an x from the first two terms and a 1 from the last two terms: Now factor the common binomial (5x 1): x(5x 1) 1(5x 1) (5x 1)(x 1) T his is the answer. Solve Real-World Problems Using Polynomial Equations Now that we know most of the factoring strategies for quadratic polynomials, we can apply these methods to solving real world problems. Example 7 One leg of a right triangle is 3 feet longer than the other leg. The hypotenuse is 15 feet. Find the dimensions of the triangle. Let x = the length of the short leg of the triangle; then the other leg will measure x

57 Chapter 1. Polynomials Use the Pythagorean Theorem: a 2 + b 2 = c 2, where a and b are the lengths of the legs and c is the length of the hypotenuse. When we substitute the values from the diagram, we get x 2 + (x + 3) 2 = In order to solve this equation, we need to get the polynomial in standard form. We must first distribute, collect like terms and rewrite in the form polynomial = 0. x 2 + x 2 + 6x + 9 = 225 2x 2 + 6x + 9 = 225 2x 2 + 6x 216 = 0 Factor out the common monomial: 2(x 2 + 3x 108) = 0 To factor the trinomial inside the parentheses, we need two numbers that multiply to -108 and add to 3. It would take a long time to go through all the options, so let s start by trying some of the bigger factors: 108 = 12 9 and = = 12 ( 9) and 12 + ( 9) = 3 T his is the correct choice. We factor the expression as 2(x 9)(x + 12) = 0. Set each term equal to zero and solve: x 9 = 0 x + 12 = 0 or x = 9 x = 12 It makes no sense to have a negative answer for the length of a side of the triangle, so the answer must be x = 9. That means the short leg is 9 feet and the long leg is 12 feet. Check: = = 225 = 15 2, so the answer checks. Example 8 The product of two positive numbers is 60. Find the two numbers if one numbers is 4 more than the other. Let x = one of the numbers; then x + 4 is the other number. The product of these two numbers is 60, so we can write the equation x(x + 4) = 60. In order to solve we must write the polynomial in standard form. Distribute, collect like terms and rewrite: x 2 + 4x = 60 x 2 + 4x 60 = 0 Factor by finding two numbers that multiply to -60 and add to 4. List some numbers that multiply to -60: 55

58 1.7. Factoring Polynomials Completely 60 = 4 15 and = = 4 ( 15) and 4 + ( 15) = = 5 12 and = 7 60 = 5 ( 12) and 5 + ( 12) = 7 60 = 6 10 and = 4 T his is the correct choice. 60 = 6 ( 10) and 6 + ( 10) = 4 The expression factors as (x + 10)(x 6) = 0. Set each term equal to zero and solve: x + 10 = 0 x 6 = 0 x = 10 x = 6 or Since we are looking for positive numbers, the answer must be x = 6. One number is 6, and the other number is 10. Check: 6 10 = 60, so the answer checks. Example 9 A rectangle has sides of length x + 5 and x 3. What is x if the area of the rectangle is 48? Make a sketch of this situation: Using the formula Area = length width, we have (x + 5)(x 3) = 48. In order to solve, we must write the polynomial in standard form. Distribute, collect like terms and rewrite: x 2 + 2x 15 = 48 x 2 + 2x 63 = 0 Factor by finding two numbers that multiply to -63 and add to 2. List some numbers that multiply to -63: 63 = 7 9 and = 2 T his is the correct choice. 63 = 7 ( 9) and 7 + ( 9) = 2 56